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Nastasia [14]
3 years ago
15

If a calibration curve of absorbance vs procaine hydrochloride concentration (mol L') is linear with the equation y = 240000x fo

r the derivatised procaine, what is the value of the molar absorptivity coefficient (Emolar)? Assume that a cuvette with a pathlength of 3 cm was used. (1 mark) Y=240000 Emolar = _ L molo
Physics
1 answer:
coldgirl [10]3 years ago
3 0

Answer:

8000 L/ mol cm is the value of the molar absorptivity coefficient.

Explanation:

The slope of the absorbance- concentration graph is equal to the product of path length and  molar absorptivity coefficient.

\frac{dA}{dc}=\epsilon l

\frac{dA}{dc}= Rate of change of absorbance with respect to concentration

\epsilon = molar absorptivity coefficient

l = Path-length

Given the slope of graph of absorbance vs procaine hydrochloride concentration.

y = 240000 x (linear equation)

Path length of cuvette = 3 cm

A=240000 c

A=\epsilon c\times l=(\epsilon l)\times c

240000=\epsilon \times 3

\epsilon =\frac{24000}{3}=8000

The units of \epsilon = 8000 l/mol cm

8000 L/ mol cm is the value of the molar absorptivity coefficient.

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GrogVix [38]

Answer:

 x =  0.176 m

Explanation:

For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.

Let's use trigonometry to decompose the tension

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      T_{y} = T sin 60

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we apply the equation

       ∑ τ = 0

       -W L / 2 - w x + T_{y} L = 0

 

the length of the bar is L = 6m

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let's calculate

let's use the maximum tension that resists the cable T = 900 N

             x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)

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5 0
3 years ago
Using Newton's 2nd Law of Motion Formula (F=MA) answer the following.
vladimir1956 [14]

Explanation:

1200 is your answer for this question

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If the cart has potential energy of 5,000 J and kinetic energy of 2,750 J. How much mechanical energy does the cart have?
Dahasolnce [82]

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Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.
olga2289 [7]

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Ratio of kilomoles oxygen to kilomole nitrogen is:

n^{*} = \frac{3.946\,kmol}{2.632\,kmol}

n^{*}= 1.499

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

NO_{1.499}

8 0
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